Exercises — Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²)

WHAT notice karo: lift height $0.80\text{m}$ Earth ke radius $6.371\times 10^6\text{m}$ ke comparison mein bahut chhoti hai. Itni chhoti range mein gravity essentially constant hai, isliye hum flat-Earth formula $U=mgh$ use karte hain, $-GMm/r$ nahi. WHY $mgh$: general well $-GMm/r$ curve karti hai, lekin kuch metres pe zoom in karo toh curve seedhi dikhti hai — ek constant downward pull $mg$. Yahi assumption $mgh$ ke peeche hai. $\Delta U=mgh=3\times 9.8\times 0.80=23.52\text{J}$ Positive hai kyunki humne gravity ke against uthaaya, energy bank ki.

WHAT: $G,M,m,r$ sab positive hain, aur aage minus sign hai, isliye $U(r)<0$ — negative. WHY negative: reference $U=0$ ko $r=\infty$ pe choose kiya jaata hai (infinitely door, koi interaction nahi). Koi bhi real orbit infinity se closer hoti hai, gravitational "well" ke andar baithti hai, isliye zero se neeche hoti hai. $U=0$ ka matlab hoga ke satellite ne barely infinite distance tak escape kiya aur kuch bacha nahi. Answer: negative; $U=0$ infinitely door hone se correspond karta hai.

WHAT: $U=\frac{1}{2}k{x}^2$ $x^2$ pe depend karta hai, aur $(-0.05{)}^2=(+0.05{)}^2$. WHY: squaring sign ko mita deta hai, isliye equal amounts se stretch aur compression same energy store karte hain. Spring ko direction ki parwaah nahi, sirf natural length se kitna door hai. Answer: $0.4\text{J}$ — bilkul same. The implied $k$: $0.4=\frac{1}{2}k(0.05{)}^2=\frac{1}{2}k(0.0025)$ solve karo, $k=0.4/(0.00125)=320\text{N/m}$ milta hai.

Step 1 — store: $U=\frac{1}{2}k{x}^2=\frac{1}{2}(500)(0.08{)}^2=\frac{1}{2}(500)(0.0064)=1.6\text{J}$. Step 2 — convert: frictionless hai, isliye saari PE kinetic energy ban jaati hai: $\frac{1}{2}m{v}^2=1.6$. WHY conservation applies: koi non-conservative force energy nahi le jaati (dekho Conservation of mechanical energy). $v=\sqrt{\frac{2\times 1.6}{0.20}}=\sqrt{16}=4\text{m/s}$ Answer: $U=1.6\text{J}$, $v=4\text{m/s}$.

WHAT: poori well depth pay karni hogi: $\Delta U=U_{\infty }-U_R=0-\left(-\frac{GMm}{R}\right)=\frac{GMm}{R}$. $\Delta U=\frac{(6.674\times 10^{-11})(5.972\times 10^{24})(1200)}{6.371\times 10^6}\approx 7.506\times 10^{10}\text{J}$ WHY this is the full well: infinity pe interaction khatam ho jaata hai ($U=0$); surface pe $U$ sabse zyada negative hoti hai, isliye bahar nikalte waqt exactly $GMm/R$ cost hoti hai. Isliye Escape velocity $\frac{1}{2}m{v}_e^2=GMm/R$ satisfy karti hai.

WHAT: master rule $F=-\frac{dU}{dx}$ use karo Force from potential — F = -dU/dx se. $\frac{dU}{dx}=8x-3\Rightarrow F=-(8x-3)=3-8x$ At $x=2$: $F=3-16=-13\text{N}$. WHY the minus: force energy landscape mein downhill ki taraf point karta hai, lower $U$ ki taraf. Negative $F$ matlab yeh $-x$ ki taraf push karta hai.

(a) Flat-Earth: $\Delta U=mgh=10\times 9.8\times 200=19600\text{J}$. (b) Exact: $\Delta U=-GMm\left(\frac{1}{R+h}-\frac{1}{R}\right)$ $R=6.371\times 10^6$, $h=200$ ke saath: $\Delta U\approx 1.9636\times 10^4\text{J}\approx 19636\text{J}$ WHAT this shows: dono mein sirf lagbhag $0.18\%$ ka fark hai — $200\text{m}$ ki lift ke liye negligible. WHY they agree: $h\ll R$ ke liye, Taylor-expand karne se $\Delta U\approx GMmh/R^2=(GM/R^2)mh=mgh$ milta hai, kyunki $g=GM/R^2$. $mgh$ literally exact expression ka pehla term hai.

Neeche wali figure kya dikhati hai: blue curve poora well $U(r)=-GMm/r$ hai, jo upar ki taraf bend karta hai $U=0$ (dotted yellow line) ki taraf jaise $r\to \infty$. Pink straight line us curve ka tangent hai surface $r=R$ pe — iska slope local $mgh$ rule hai. $200\text{m}$ jaise tiny height pe blue curve aur pink tangent visually alag nahi dikhte, isliye kyun $mgh$ ground ke paas kaam karta hai: aap ek gently curving well ke straight-line approximation pe chal rahe ho.

WHAT: force zero hoti hai jahan $\frac{dU}{dx}=0$ ho (landscape mein flat spot). $\frac{dU}{dx}=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3)=0\Rightarrow x=1,3$ Second derivative se classify karo $\frac{d^{2}U}{d{x}^2}=6x-12$:

• $x=1$ pe: $6(1)-12=-6<0$ → $U$ ka maximum → unstable (ball roll away karti hai).
• $x=3$ pe: $6(3)-12=+6>0$ → $U$ ka minimum → stable (ball wapas settle ho jaati hai). WHY: valley bottom ($U$ min) nudged bead ko wapas push karta hai; hilltop ($U$ max) use aur door push karta hai.

Neeche wali figure kya dikhati hai: blue curve $U(x)$ ko $x$ ke across plot karta hai. Pink dot $x=1$ pe local hilltop pe hai — bead ko kisi bhi taraf nudge karo aur $U$ drop karta hai, toh woh door accelerate karta hai: unstable. Yellow dot $x=3$ pe local valley mein hai — nudge karo aur $U$ dono taraf rise karta hai, toh force use wapas push karta hai: stable. Picture concrete karta hai kyun second derivative ka sign (neeche curve karna vs upar curve karna) exactly stability test hai.

Stored PE: $U=\frac{1}{2}(300)(0.10{)}^2=1.5\text{J}$. Momentum: rest se shuru hota hai, isliye total momentum zero rehta hai: $m_1{v}_1=m_2{v}_2$, yaani $0.30v_1=0.60v_2\Rightarrow v_1=2v_2$. Energy: $\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=1.5$. $v_1=2v_2$ substitute karo: $\frac{1}{2}(0.30)(2v_2{)}^2+\frac{1}{2}(0.60)v_2^2=0.60v_2^2+0.30v_2^2=0.90v_2^2=1.5$ $v_2=\sqrt{1.5/0.90}=\sqrt{5/3}\approx 1.291\text{m/s},v_1=2v_2\approx 2.582\text{m/s}$ WHY dono laws: energy akele do unknowns ke liye ek equation deta hai; momentum conservation doosra deta hai. Halka block zyada tezi se uda jaata hai.

WHAT: maximum compression pe ball momentarily rest mein hai, isliye saara gravitational PE lost spring PE stored ke barabar hai. Ball total height $h+x$ girti hai (drop plus compression). Energy balance ($U=0$ spring ki top pe): $mg(h+x)=\frac{1}{2}k{x}^2$ $0.50\times 9.8\times (1.2+x)=\frac{1}{2}(800)x^2$ $4.9(1.2+x)=400x^2\Rightarrow 400x^2-4.9x-5.88=0$ Quadratic solve karo: $x=\frac{4.9\pm \sqrt{4.9^2+4(400)(5.88)}}{2(400)}=\frac{4.9\pm \sqrt{9432.01}}{800}$ $x\approx \frac{4.9\pm 97.12}{800}\Rightarrow x\approx +0.1275\text{m}\text{ or }x\approx -0.1152\text{m}$ WHY negative root discard karo: $x$ ek compression distance hai, ek physical length jitna spring neeche squeeze hoti hai — yeh positive hona chahiye. Negative root ka matlab hoga spring upar stretch hoti hai, lekin girti ball spring ko sirf neeche push kar sakti hai, isliye woh root unphysical hai aur hum $x\approx +0.128\text{m}$ rakhte hain.

Escape speed check: $v_e=\sqrt{2GM/R}=\sqrt{2(6.674\times 10^{-11})(5.972\times 10^{24})/6.371\times 10^6}\approx 11186\text{m/s}$. Kyunki $8000<11186$, yeh escape nahi karta — theek hai, ek finite $r_{\max }$ exist karta hai. Energy conservation (KE + PE conserved; $r_{\max }$ pe speed = 0): $\frac{1}{2}m{v}_0^2-\frac{GMm}{R}=0-\frac{GMm}{r_{\max }}$ Cancel $m$, $r_{\max }$ ke liye solve karo: $\frac{GM}{r_{\max }}=\frac{GM}{R}-\frac{1}{2}{v}_0^2\Rightarrow r_{\max }=\frac{GM}{\frac{GM}{R}-\frac{1}{2}{v}_0^2}$ Numbers: $GM=3.986\times 10^{14}$, $GM/R=6.256\times 10^7$, $\frac{1}{2}{v}_0^2=3.2\times 10^7$. $r_{\max }=\frac{3.986\times 10^{14}}{6.256\times 10^7-3.2\times 10^7}=\frac{3.986\times 10^{14}}{3.056\times 10^7}\approx 1.304\times 10^7\text{m}$ WHY $-GMm/r$ use karo, $mgh$ nahi: rocket hazaron km chadhta hai, jahan gravity noticeably weak hoti hai — $mgh$ ka constant-$g$ assumption toot jaata hai. Answer: $r_{\max }\approx 1.30\times 10^7\text{m}$ (lagbhag $2.05R$).

WHAT "leaving" matlab: ball tab jaati hai jab spring ka contact force zero ho jaata hai. Spring sirf push kar sakti hai (natural length pe pahunchne ke baad ball ko pull nahi kar sakti). Isliye contact exactly natural length pe khatam hota hai, $x_{\text{leave}}=0$. WHY equilibrium point pe nahi: ek aam guess yeh hoti hai ke spring force gravity balance karta hai ($kx=mg\Rightarrow x=mg/k=4.9/800\approx 0.006\text{m}$). Lekin wahan spring abhi bhi compressed aur push kar rahi hai; ball sirf tab separate hoti hai jab spring aur push nahi kar sakti, jo natural length pe hota hai. Uske aage spring pull karti — non-attached ball ke liye impossible. Answer: $x_{\text{leave}}=0$ (natural length). Equilibrium compression $mg/k\approx 0.0061\text{m}$ woh jagah hai jahan net force zero hai, lekin ball chalti rehti hai aur $x=0$ pe jaati hai.

$r\to \infty$: $U\to 0^{-}$. Do masses infinitely door hain, effectively non-interacting. Yeh chosen zero reference hai — clean aur physical. $r\to 0^{+}$: $U\to -\infty$. Formula ek infinitely deep well predict karta hai. KAHAN break karta hai: real objects finite size ke hote hain. Aap do centres ko $r=0$ tak nahi la sakte kyunki aap pehle surface se takraate ho: radius $R_p$ wale planet ke liye, point-mass law $-GMm/r$ sirf body ke bahar valid hai ($r\ge R_p$). Jab aap andar jaate ho, toh aapke neeche ka mass sab andar nahi kheenchta aur Newton's shell theorem poora law badal deta hai — $U$ asli centre pe actually finite aur smooth rehta hai, yeh blow up nahi karta. Isके upar, extreme density pe Newtonian gravity itself general relativity ko rasta de deti hai. Isliye $r\to 0$ ek purely mathematical limit hai ek equation ka jo apne domain ke bahar use hoti hai, ordinary bodies ke liye koi reachable state nahi. WHY limits track karo: yeh dikhate hain ke "bottomless well" us assumption ka artefact hai ke mass ek point hai. $U$ ka real, physical minimum surface $r=R$ pe hota hai, $U=-GMm/R$ deta hai — woh finite depth jo hum actually escape karne ke liye pay karte hain (jaisa L2.2 mein).

WHAT sign matlab:

• $E<0$ → bound: KE poori well depth pay nahi kar sakti, isliye kisi $r_{\max }$ pe speed zero ho jaati hai aur woh wapas girta hai.
• $E=0$ → marginal: zero speed ke saath $r=\infty$ tak pahunchta hai (exactly escape).
• $E>0$ → unbound: infinity pe leftover KE, escape karta hai aur chalata rehta hai. Test: $\frac{1}{2}(1)(11186{)}^2=6.256\times 10^7\text{J}$; $\frac{GMm}{R}=\frac{3.986\times 10^{14}}{6.371\times 10^6}=6.256\times 10^7\text{J}$. $E=6.256\times 10^7-6.256\times 10^7\approx 0$ WHY: $v=11186\text{m/s}$ hi escape speed hai, isliye $E=0$ — boundary case. Thoda tez ⇒ $E>0$ ⇒ escape; thoda slow ⇒ $E<0$ ⇒ bound. Yeh ek sign orbits ki poori kahani hai.

$mgh$ sirf $h\ll R$ ke liye valid hai ::: over large climbs use $-GMm/r$. Spring energy displacement ko square karta hai ::: $\frac{1}{2}k{x}^2$, kabhi $\frac{1}{2}kx$ nahi. Equal spring force ⇒ equal momentum, equal speed nahi ::: halka block tez jaata hai. Force $U$ ke flat spots pe zero hoti hai; stable = minimum, unstable = maximum ::: $d^{2}U/d{x}^2$ ka sign use karo. Total energy ka sign fate decide karta hai ::: $E<0$ bound, $E=0$ escape, $E>0$ unbound. Ball spring ko natural length pe chodti hai ::: force-balance point pe nahi.